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2 years ago

Tyler made a scale drawing of his apartment. The scale is 1 millimeter: 2 meters. The living room is 6 meters long in

Mathematics

1 answer:

mojhsa [17]2 years ago

4 0

I think the answer would be 2years

You might be interested in

Two less than 3/2 of a number (x) is more than 5 1/2

garik1379 [7]

Answer:

3/2x-2>5 1/2

Step-by-step explanation:

Two less than a number means we are subtracting 2, so we can write it into the equation as -2.

Since we dont know the value of x, we can just write 3/2 of x by multiplying 3/2 by x. So 3/2x. Also, since we are subtracting two from 3/2x, -2 comes after 3/2x.

Now, since it says that it is more than 5 1/2, we could use the greater than sign (>), to express that. So >5 1/2

Our final expression is 3/2x-2>5 1/2.

Hope this helps!

3 0

3 years ago

You buy hamburgers at a fast food restaurant. A hamburger costs $0.49. You must have at most $3 to spend. Write an inequality fo

KIM [24]

Answer:

x ≤ 6.12

Step-by-step explanation:

If the cost of 1 hamburger in a restaurant is $0.49, the amount of hamburger $3 can buy is expressed as shown .

$0.49 = 1

$3 ≤ x (since you have maximum of $3 to spend. It cant be more than that)

cross multiply to find the expression for x.

$0.49 * x ≤ $3 * 1

$0.49x ≤ $3

Divide both sides by $0.49

$0.49x/$0.49 ≤ $3/$0.49

x ≤ $3/$0.49

x ≤ 6.12

The number of hamburgers you can buy is represented by the inequality x ≤ 6.12

4 0

2 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

3 years ago

I for really need help with this one and a couple of others HELPPPP

Keith_Richards [23]

Check the picture below.

8 0

3 years ago

Let F be the event that a student is enrolled in a finance course, and let S be the event that a student is enrolled in a statis

Inessa [10]

Answer:

0.375

Step-by-step explanation:

Given that F is the event that a student is enrolled in a finance course, and let S be the event that a student is enrolled in a statistics course.

It is known that 40% of all students are enrolled in a finance course and 35% of all students are enrolled in statistics. Included in these numbers are 15% who are enrolled in both statistics and finance.

P(F) = 0.40: P(S) = 0.35: P(FS) = 0.15

A student is randomly selected, and it is found that the student is enrolled in finance.

Probability that this student is also enrolled in statistics/ student is enrolled in finance.

= $\frac{P(FS)}{P(F)} \\=\frac{0.15}{0.40} \\=0.375$

(Using conditional probability formula)

5 0

3 years ago